Question

If the $10th$ term of an $A.P.$ is $\frac{1}{20}$ and the $20th$ term is $\frac{1}{30}$, then find the $200th$ term.

Answer 1

Given that,

$a_{10}=\frac{1}{20}$

$a_{20}=\frac{1}{30}$

To find:- $a_{200}=?$

Solution:-

As we know that the $n^{th}$ term of an A.P. is given by-

$a_n=a+(n-1)d$

Here $a$ and $d$ are first term and common difference of A.P. respectively.

Therefore,

$a+9d=\frac{1}{20}.....\left(1\right)$

$a+19d=\frac{1}{30}$

$\Rightarrow (a+9d)+10d=\frac{1}{30}$

$\Rightarrow \frac{1}{20}+10d=\frac{1}{30}\left[\text{From}\left(1\right)\right]$

$\Rightarrow 10d=\frac{1}{30}-\frac{1}{20}$

$\Rightarrow d=-\frac{1}{600}$

Substituting the value of $d$ in ${eq}^n\left(1\right)$, we have

$a+9\left(\frac{-1}{600}\right)=\frac{1}{20}$

$\Rightarrow a=\frac{1}{20}+\frac{9}{600}=\frac{39}{600}$

$\therefore a_{200}=a+199d$

$\Rightarrow a_{200}=\frac{39}{600}+199(-\frac{1}{600})$

$\Rightarrow a_{200}=-\frac{160}{600}=-\frac{4}{15}$

Hence the correct answer is $-\frac{4}{15}$.